It is the same kind of results as [1]} adapted to this difference context. Let \(H\)  be the smallest algebraic group of upper bloc triangular matrices such that \(B\in H(\mathbf {k})\) . It is an algebraic group whose elements are of the form \(\begin{pmatrix} H_1 & H_2 \\ 0 & H_3 \end{pmatrix}\) . We denote by \(H_d\)  the group whose matrices are of the form \(\begin{pmatrix} H_1 & 0\\ 0 & H_3 \end{pmatrix} \)  where \(\begin{pmatrix} H_1 & H_2\\ 0 & H_3 \end{pmatrix}\in H \)  for a matrix \(H_2\) . The group \(H_d\)  is a block diagonal algebraic group and is the smallest algebraic group
such that \(D\in H_d (\mathbf {k})\) . Since \([D]\)  is in reduced form, we have \(G_d= H_d\) .
Let \(R = \begin{pmatrix} R_1 & R_2 \\ 0 & R_3 \end{pmatrix}\in H(\mathbf {k})\)  be such that the system \(\left[\phi (R)BR^{-1}\right]\)  is in reduced form. Let \(R_d :=\begin{pmatrix} R_1 & 0 \\ 0 & R_3 \end{pmatrix} \in H_d(\mathbf {k})=G_d(\mathbf {k})\) .
Let \(G_u\)  be the subgroup of \(G\)  whose elements are of the form \(\begin{pmatrix}\mathrm {Id}_m & \star \\ 0 & \mathrm {Id}_{n-m}\end{pmatrix}\) . We have \(G_d= G/G_u\) .
Therefore, we find that there exists \(Q_2\) , such that \(Q:= \begin{pmatrix}R_1 & Q_2 \\ 0 & R_3\end{pmatrix}\in G(\mathbf {k})\) . Therefore \(Q^{-1} = \begin{pmatrix}R_1^{-1} &\star \\ 0 & R_3^{-1}\end{pmatrix}\in G(\mathbf {k})\) . A gauge transformation of \(\phi (R)BR^{-1}\in G(\mathbf {k})\)  by the matrix \(Q^{-1}\in G(\mathbf {k})\)  gives an element of \(G(\mathbf {k})\) . Hence \(\left[\phi (T)BT^{-1}\right]\)  is in reduced form where \(T:= Q^{-1}R = \begin{pmatrix}\mathrm {Id}_m & \star \\ 0 & \mathrm {Id}_{n-m}\end{pmatrix}\) .
